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Why isn't the Weierstrass function $\sum_{n=0}^\infty a^n \cos(b^n\pi x)$ differentiable?

Nothing is preventing us from taking the derivative of any finite partial sum of this series. This is a trigonometric polynomial and it has derivatives of all orders. However, this infinite sum ...
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Finding $\displaystyle \sum \limits _{n=1}^{\infty}\frac{(-1)^n (H_{2n}-H_{n})}{n(2)^n \binom{2n}{n}}$

Clear["Global`*"] f[n_] := (-1)^ n (HarmonicNumber[2 n] - HarmonicNumber[n])/(n*2^n*Binomial[2 n, n]) The terms of the sum converge to zero rapidly <...
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